1020. Tree Traversals (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

分析:根据题目所给的后序遍历序列和中序遍历序列，构建二叉树，再输出该二叉树的前序遍历序列。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

const int maxn=40;
int post[maxn];
int in[maxn];

struct node
{
    int data;
    node *lchild,*rchild;
};

node * creat(int postL,int postR,int inL,int inR)
{
    if(postL>postR) return NULL;
    node * root=new node;
    root->data=post[postR];
    int index;
    for(index=inL;index<=inR;index++)
    {
        if(in[index]==post[postR])
        {
            break;
        }
    }
    int numLeft=index-inL;
    root->lchild=creat(postL,postL+numLeft-1,inL,index-1);
    root->rchild=creat(postL+numLeft,postR-1,index+1,inR);
    return root;
}

int first_flag=0;

void layerOrder(node * root)
{
    queue<node *> ans;
    ans.push(root);
    while(!ans.empty())
    {
        node * tmp=ans.front();
        ans.pop();
        if(first_flag!=0)
        {
            cout<<" ";
        }
        cout<<tmp->data;
        first_flag=1;
        if(tmp->lchild!=NULL) ans.push(tmp->lchild);
        if(tmp->rchild!=NULL) ans.push(tmp->rchild);
    }
}

int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>post[i];
    }
    for(int i=0;i<n;i++)
    {
        cin>>in[i];
    }
    node * root;
    root=creat(0,n-1,0,n-1);
    layerOrder(root);
    return 0;
}

[二叉树建树] 后序遍历与中序遍历建立二叉树

1020. Tree Traversals (25)

推荐阅读